The sum of the series frac{1}{1} + frac{1}{1 | vedclass

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(D) The $n$-th term of the series is given by $t_n = \frac{1}{\sum_{k=1}^{n} k} = \frac{1}{\frac{n(n + 1)}{2}} = 2 \left[ \frac{1}{n} - \frac{1}{n + 1

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$\frac{2(n + 1)}{n + 2}$

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(D) The $n$-th term of the series is given by $t_n = \frac{1}{\sum_{k=1}^{n} k} = \frac{1}{\frac{n(n + 1)}{2}} = 2 \left[ \frac{1}{n} - \frac{1}{n + 1} \right]$.To find the sum up to $(n + 1)$ terms,we calculate $S_{n+1} = \sum_{k=1}^{n+1} t_k$.$S_{n+1} = 2 \left[ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dots + \left( \frac{1}{n + 1} - \frac{1}{n + 2} \right) \right]$.This is a telescoping series,so $S_{n+1} = 2 \left[ 1 - \frac{1}{n + 2} \right]$.$S_{n+1} = 2 \left( \frac{n + 2 - 1}{n + 2} \right) = \frac{2(n + 1)}{n + 2}$.

The sum of the series $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ up to $(n + 1)$ terms is equal to:

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